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User talk:Hyp cos/OCF vs Array Notation p2
Beating weakly compact cardinal I think \(K\) = \(M(1;0;0)\), and \(\{1\{1,,1\{1\{1,,1,,2^{,,}\}2\}2^{,,}\}2\}\) reaches that. Googleaarex (talk) 00:31, August 3, 2017 (UTC) : Are you sure? In my analysis, K = {1{1,,1,,2,,}2}, and Ψ(Ψ(Ψ(1,1)+λ0(Ψ(λ0(Ψ(λ0,1)),1)),1)) (in lambda notation) reaches the limit of pDAN. Nishada 01:06, August 3, 2017 (UTC) : '''EDIT: '''I meant to say that Ψ(Ψ(Ψ(1,1)+λ0(Ψ(λ0,1)),1)) reaches the limit of PDAN. Nishada 02:31, August 3, 2017 (UTC) : '''EDIT 2: '''I reanalyzed and found out that the limit of PDAN is Ψ(Ψ(Ψλ0(0),1)). Nishada 19:01, August 3, 2017 (UTC) :: Maybe Lambda Notation beats all of SAN extensions? Googleaarex (talk) 20:04, August 3, 2017 (UTC) : Hyp cos updated the analysis and found out \(K\) > \(M(1;0;0)\), so I will be using your analysis. Googleaarex (talk) 11:41, August 4, 2017 (UTC) Analyze with lambda notation next! Nishada 16:56, August 5, 2017 (UTC) :Lambda notation might be ill-defined. Googleaarex (talk) 18:10, August 5, 2017 (UTC) :: Why? Nishada 18:44, August 5, 2017 (UTC) ::: Wow, now you are calling each other's ordinal notation ill-defined? -- ☁ I want more ⛅ 03:05, August 6, 2017 (UTC) Guess about collapsing stable ordinal or total indescribables If s(x,x{1{1`,,2,,}2}2) - the growth rate limit in KP+\(\Pi_n\)-reflection or n-admissible or first \(\Pi^1_n\)-indescribable (not fitting Taranovsky's correspondence) What comparisons (not fitting Taranovsky's correspondence) will be made with KP+\(\Pi_\omega\)-reflection or KP+\(\Pi^1_0\)-reflection or 1-stable or \(\Pi^2_0\)-indescribable (or 1-\(\Pi^1_0\)-indescribable fitting Stegert correspondence) - Small Stegert ordinal? \(\Pi^3_0\)-indescribable (or 2-\(\Pi^1_0\)-indescribable fitting Stegert correspondence)? \(\Pi^n_0\)-indescribable (or n-\(\Pi^1_0\)-indescribable fitting Stegert correspondence)? \(\Pi^n_m\)-indescribable (or\(\omega\)-\(\Pi^1_0\)-indescribable fitting Stegert correspondence)? \(\alpha\)-\(\Pi^1_0\)-indescribable fitting Stegert correspondence - Large Stegert ordinal? Why fitting Taranovsky's correspondence and another guess is different?Scorcher007 (talk) 05:41, January 4, 2018 (UTC) :I don't know, because KP + \(\Pi_n\)-reflection reaches a very special point - s(x,x{1,,,3}2). And I can't make a guess. {hyp/^,cos} (talk) 09:30, January 4, 2018 (UTC) stable ordinal comparison Assuming that DAN ordinal is PTO of П12-CA0 then the collapse of stable ordinal will be stronger My intuitive analysis is based on Stegert, Taranovsky, Feferman and Madore paper. First, we introduce some notation: Ξ0 = ω1CK = 1st 1-admissible (2.1) ~card.equiv. = ω1 = 1st uncountable = 1st 1-regular Ξ02 = ω2CK = 2nd 1-admissible ~card.equiv. = ω2 = 2nd uncountable = 2nd 1-regular Ξ1 = П3-reflection = 2-admissible (2.6) ~card.equiv. = П11-indescribable = 2-regular = weakly copmact Ξ2 = П4-reflection = 3-admissible ~card.equiv. = П12-indescribable = 3-regular Ξn = Пn+2-reflection = n+1-admissible ~card.equiv. = П1n-indescribable = n+1-regular collapse(Ξn) is limit of PDAN Ξω = Пω-reflection = П10-reflection = ω-admissible = (+1)-stable (2.7) ~card.equiv. = 1-П10-indescribable = П1ω-indescribable = П20-indescribable = ω-regular Сollapse(Ξω) will be stronger. Because as I understand we can do something like that Сollapse(Ξω+1) = Сollapse(Ξn+1) Сollapse(εΞω+1) is small Streget ordinal (1.25) After: Ξω×2 = ω×2-admissible ordinal = (+2)-stable ~card.equiv. = 2-П1ω-indescribable = П30-indescribable = ω×2-regular Ξω×n = ω×n-admissible ordinal = (+n)-stable ~card.equiv. = n-П1ω-indescribable = Пnm-indescribable = ω×n-regular Ξω2 = ω2-admissible ordinal = (+ω)-stable ~card.equiv. = ω-П10-indescribable = ω2-regular Ξε0 = ε0-admissible ordinal = (+ε0)-stable ~card.equiv. = ε0-П10-indescribable = ε0-regular Ξ[Ξ0] = 1st 1-admissible-admissible ordinal = (+)-stable (2.8) ~card.equiv. = ω1CK-П10-indescribable Ξ[Ξ02] = 2nd 1-admissible-admissible ordinal = (++)-stable (2.10) ~card.equiv. = ω2CK-П10-indescribable ΞI = recursively inaccessible 1-admissible-admissible ordinal = inaccessibly-stable (2.11) ~card.equiv. = I-П10-indescribable ΞM = recursively Mahlo 1-admissible-admissible ordinal = Mahlo-stable (2.12) ~card.equiv. = M-П10-indescribable Ξ[Ξ1] = 2-admissible-admissible ordinal = 2-admissible-stable ~card.equiv. = Ξ1-П10-indescribable Ξ[Ξω] = ω-admissible-admissible ordinal = double (+1)-stable ordinal (2.13) ~card.equiv. = Ξω-П10-indescribable After introduce some another notation: ϒ{0} = ω ϒ{1} = Ξω = (+1)-stable ϒ{2} = Ξ[Ξω] = double (+1)-stable ϒ{3} = Ξ[Ξ[Ξω]] = triple (+1)-stable ϒ{n} = n-le (+1)-stable ordinal ϒ{ω} = ω-le (+1)-stable ordinal = 1st nonprojectible ordinal = strongly admissible = П12-comprehension (2.15) Сollapse(ϒ{ω}) is PTO of П12-CA0 and if what I assumed at the beginning is true, then Сollapse(ϒ{ω}) is limit of DAN. Then it turns out that Сollapse(Ξ[Ξω]) or double (+1)-stable ordinal is limit of SDAN. Сollapse(εϒ{ω}+1) is large Streget ordinal (1.26) and probably is PTO of П12-CA-TI and bigger that DAN or equals Dropping hydra ordinal. We can continue: ϒ{α} = ϒ{ϒ{ϒ{ϒ{...}}}} may be (2.16)-ordinal-ordinal and collapse(ϒ{α}) is PTO of П12-TR0 and much bigger that DAN ordinal. I can also assume that the collapse of (2.17)-ordinal is PTO of Z2 Scorcher007 (talk) 04:16, January 18, 2018 (UTC) I was wrong, П12-CA0 much stronger than ϒ{ω}. According to Rathjen Maybe it looks like this ϒ{1} = Ξω = (+1)-stable ~card.equiv. = 1-П10-indescribable ϒ{2} = Ξ[Ξω] = double (+1)-stable ~card.equiv. = 1-П10-indescribable-П10-indescribable ϒ{3} = Ξ[Ξ[Ξω]] = triple (+1)-stable ~card.equiv. = 1-П10-indescribable-П10-indescribable-П10-indescribable ϒ{n} = n-le (+1)-stable ordinal ψ(ϒ{n}) - SDAN limit ϒ{ω} = ω-le (+1)-stable ordinal = nonprojectible-stable ordinal (2.14) ~card.equiv. = subtle ψ(εϒ{ω}+1) - large Stegert ordinal ϒ{ϒ{ϒ{ϒ{ϒ{...}}}}} = (unbounded in any stable ordinal)-ordinal = 1st nonprojectible ordinal = strongly admissible ordinal (2.15)~card.equiv. = Erdos ψ(ϒ{ϒ{ϒ{ϒ{ϒ{...}}}}}) = П12-comprehension, DAN limit Scorcher007 (talk) 09:30, January 20, 2018 (UTC) And again, correction. It seems П12-comprehension is even stronger: Ϛ{0} = Ξ0ω = ωωCK = ω-th 1-admissible ordinal = П11-comprehension ordinal ~card.equiv. = ωω = ω-th uncountable = ω-th 1-regular ψ(Ξ0ω) - meAN limit Ϛ{1} = Ξω = Пω-reflection = П10-reflection = ω-admissible ordinal = (+1)-stable ~card.equiv. = 1-П10-indescribable = П1ω-indescribable = П20-indescribable = ω-regular ψ(Ξω) - PDAN limit ψ(εΞω+1) - small Stegert ordinal Ϛ{2} = ϒω = Ξ[Ξ[Ξ[Ξ[Ξ...]]]] = ω-le (+1)-stable ordinal = 1st nonprojectible-stable ordinal ~card.equiv. = subtle ψ(ϒω) - SDAN limit ψ(εϒω+1) - large Stegert ordinal ϒ[ϒω] = 1st nonprojectible-stable nonprojectible-stable ordinal Ϛ{3} = ϒ[ϒ[ϒ[ϒ[ϒ...]]]] = 1st nonprojectible-nonprojectible-stable ordinal ψ(Ϛ{3}) - TDAN limit Ϛ{ω} = (unbounded in any stable ordinal)-ordinal = 1st nonprojectible ordinal = П12-comprehension ordinal = strongly admissible ordinal ~card.equiv. = Erdos ψ(Ϛ{ω}) - DAN limit, PTO of П12-CA0 Scorcher007 (talk) 04:22, January 22, 2018 (UTC) I may need to point out that the limit of sDAN is {1,,,1,,,1...,,,1,,,2} (or {1{2,,,,2}2}), not {1,,,,3}; this limit is not similar to the limit of pDAN, which is {1{1...{1,2,,}...2,,}2,,} (or {1,,,3}). {hyp/^,cos} (talk) 05:03, January 22, 2018 (UTC) If I understand correctly: 1st part PDAN limit is {1,,1,,1,,1,,1,,1 ... 2} = {1{2,,}2} = {1{2,,,2}2} 2nd part PDAN limit is {1{1{1...{1,2,,}...2,,}2,,}2} = {1{1{1,,,3}2,,,2}2} = {1,,,3} 1st part SDAN limit is {1{1,,...2,,}2} = {1{1{1,,,1,2}2,,,2}2} = {1,,,1,2} 2nd part SDAN limit is {1,,,1,,,1,,,1,,,1,,,1 ... 2} = {1{2,,,,2}2} Which expression should be considered DAN limit: {1{2,,,...2}2} or {1,,,...2} or {1,,,...3} or {1,,,...1,2} ? Scorcher007 (talk) 13:07, January 22, 2018 (UTC) :{1,,...,,3}, {1,,...,,1,2} and {1{2,,...,,2}2} are all limit of DAN, because {1,,...,,3} (with an m-ple comma) < {1,,...,,1,2} (with an m-ple comma) < {1{2,,,...,,2}2} (with an (m+1)-ple comma) < {1,,,...,,3} (with an (m+1)-ple comma). {hyp/^,cos} (talk) 16:08, January 22, 2018 (UTC) :Then it's really hard to find patterns for comparison, even if you imagine that DAN limit is П12-comprehension. Dropping hydra allows you to make a prediction that DAN is П12-comprehension, because Kirby-Paris hydra has limit is Arithmetical comprehension, and Buchholz hydra has limit s П11-comprehension. They are associated with subsystems of Z2. But then it's hard to make assumptions about SDAN limit. It can only be said that it corresponds to some stable ordinal. :Perhaps, according to my notation: :Ϛ{1} = Ξω = 1st-stable ordinal ~ {1,,,3} :Ϛ{2} = ϒω = Ξ[Ξ[Ξ[Ξ[Ξ...]]]] = 1st nonprojectible-stable ordinal ~ {1,,,,3} :Ϛ{3} = ϒ[ϒ[ϒ[ϒ[ϒ...]]]] = 1st nonprojectible-nonprojectible-stable ordinal ~ {1,,,,,3} :e t.c. up to :Ϛ{ω} = (unbounded in any stable ordinal)-ordinal = 1st nonprojectible ordinal ~ {1,,...,,3} Scorcher007 (talk) 16:49, January 22, 2018 (UTC) :If DAN stronger than П12-comprehension, then we can assume that: :{1'2} ~ ACA0 :{1,,2} ~ ATR0 :{1,,3} ~ KPω :{1,,1,2} ~ П11-CA0 :{1,,1,,2} ~ П11-TR0 :{1,,,3} ~ Ξω = (+1)-stable or KP+Пω, where ψ(εΞω+1) - small Stegert ordinal :{1,,,4} ~ ϒω = ω-le (+1)-stable ordinal or KP+stability = nonprojectible-stable ordinal, where ψ(εϒ{ω}+1) - large Stegert ordinal :{1,,,1,2} ~ П12-CA0 :{1,,,1,,,2} ~ П12-TR0 :{1,,,,1,2} ~ П13-CA0 :{1,,,,1,,,,2} ~ П13-TR0 :{1,,,...,,,1,2} ~ Z2 Scorcher007 (talk) 02:51, February 1, 2018 (UTC) ::{1,,2} is identical to the grave accent, so s(a,b{1,,2}2) = s(a,b`2) = s(a,b{1`2}2). If {1`2} corresponds to ACA0, then so does {1,,2}. In addition, what corresponds to ATR0 (\(\Gamma_0\)) is {1`1`1`2}. {hyp/^,cos} (talk) 04:48, February 1, 2018 (UTC) :::By the way, it is unclear what to consider as the SAN equal limit, below П11-CA0 ? FGH(3) ~ SGH(ε0) FGH(ω+1) ~ SGH(Г0) FGH(ωω) ~ SGH(Ωω) ~ {2} FGH(ε0) ~ SGH(εΩ+1) ~ {1`2} FGH(Г0) ~ SGH(ГΩ+1) ~ {1`1`1`2} FGH(εΩ+1) ~ SGH(εΩ2+1) ~ {1,,2} and only FGH(Ωω) ~ SGH(Ωω) ~ {1,,1,2} :::Scorcher007 (talk) 13:01, February 1, 2018 (UTC) Loader's number question Is the limit of DAN is larger than even Loader's number? (sorry for short question) ARsygo (talk) 14:34, January 16, 2019 (UTC) :No, Loader's function is larger. Limit of DAN less than П12-CA0. П12-CA0 is limit of Rathjen's OCF. Next stage: П1ω-CA0 = Z2 (Second order arithmetic). Next stage: Zω (Higher order arithmetic). Loader's function limit approximately corresponds to Higher order arithmetic. More detailed approximate comparison of this scale can be found here Scorcher007 (talk) 15:32, January 16, 2019 (UTC). : This OCF is unable to represent an inaccessible mahlo mahlo cardinal. Can you fix that?12AbBa (talk) 14:04, September 17, 2019 (UTC) :In which OCF (which section)? And what do you mean by "inaccessible mahlo mahlo cardinal"? {hyp/^,cos} (talk) 14:24, September 17, 2019 (UTC) Small adjustment to your Guess It seems to me that 1-П1-reflection can't be admissible ordinal. It must be the limit of admissible ordinals. I suppose that 1-П1-reflection is limit of level-(n<ω) +1-stable. And just 1-П2-reflection is admissible on levels 1-stable. According to Guess 3 it must be: {1{1`,,1,,2,,}2} is 1-stable {1{1`,,1,,1,2,,}2} is 1-П1-reflection {1{1`,,1,,1,,2,,}2} is 1-П2-reflection {1{1`,,1,,1,,1,2,,}2} is 1-П1-reflection on 1-П2-reflection {1{1`,,1,,1,,1,,2,,}2} is 1-П2-reflection that is 1-П1-reflection on 1-П2-reflection {1{1`,,1{1,,2,,}2,,}2} is 1-П2-reflection on 1-П2-reflection {1{1`,,1{1,,1,,2,,}2,,}2} is 1-П3-reflection {1{1`,,1{1{1,,2,,}2,,}2,,}2} is 1-П4-reflection e.t.c. Similarly to smaller equivalents: {1,,2} is П2-reflection {1,,1,2} is П1-reflection on П2-reflection {1,,1,,2} is П2-reflection that is П1-reflection on П2-reflection {1,,1,,1,2} is П1-reflection on П2-reflection that is П1-reflection on П2-reflection {1,,1,,1,,2} is П2-reflection that is П1-reflection on П2-reflection that is П1-reflection on П2-reflection {1{1,,2,,}2} is П2-reflection on П2-reflection {1{1,,1,,2,,}2} is П3-reflection {1{1{1,,2,,}2,,}2} is П4-reflection e.t.c. Scorcher007 (talk) 02:46, October 21, 2019 (UTC)